# Laplace Equation on Rectangle & Fourier Series

I am trying to solve the Laplace Equation on the rectangle. Namely, I trying to solve the following PDE problem:

Solve the equation $\Delta u = 0$ on the rectangle $R = \{(x, y): 0\leq x \leq a, \hspace{2mm} 0\leq y \leq b\}$ subject to $u(x,0) = f(x)$, $u(x, b) = 0$, $u(0, y) = 0$ and $u(a, y) = 0$.

I have separated variables and used the first three boundary conditions to obtain

$$u(x, y) = \sum_{n= 1}^{\infty} c_n \sin\left(\frac{n\pi x}{a}\right)\sinh\left(\frac{n\pi (b-y)}{a}\right)$$

Now it is the final boundary condition that seems strange. I am reading how it has been dealt with Olver's introduction book to PDEs. His argument goes as follows:

We want

$$u(x, 0) =:f(x) = \sum_{n= 1}^{\infty} c_n \sinh\left(\frac{n\pi b}{a}\right)\sin\left(\frac{n\pi x}{a}\right)$$

and this looks like the fourier sine series for $f(x)$.

So let $$b_n = \frac{2}{a} \int_0^a f(x) \sin\left(\frac{n\pi x}{a}\right)dx$$ then $c_n = \frac{b_n}{\sinh\left(\frac{n\pi b}{a}\right)}$.

But the issue I have here is that does the fact that $f$ has a fourier sine series not assume/imply that the function $f$ is odd? How is this possible - was the choice of $f$ not arbitrary? So the PDE problem could have been posed with a specific function for $f$ where $f$ is instead even and then this does not make sense anymore?

## 2 Answers

The point is that you only care about the values that $f$ takes for $0 \leq x \leq a$: you can choose the other values however you like, or not at all. All the book is doing is representing $f$ as a sine series on the interval $[0,a]$. It so happens that said sine series extends naturally to an odd function with period $2a$, but that this happens is completely irrelevant for the properties of the solution of the equation in the region considered.

• thanks, really helpful and clear – PhysicsMathsLove May 28 '18 at 15:57

You can extend $f$ to an odd function on $[-a,a]$ and expand in a sine series. If $f$ has left- and right- derivatives at $0$, then the resulting sin series will converge to $0$ at $x=0$ and at $x=a$, which may not match up with $f$ at those points. This is not unreasonable because consistency in the conditions for the PDE would require that $f(0)=f(a)=0$ because $u(0,y)=0$, $u(a,y)=0$ are also required to hold for $y$. In any event, the Fourier sine series will converge in $L^2[0,a]$ to the prescribed function.